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Thread: arithmetic progression

  1. #1
    MHF Contributor
    Sep 2008
    West Malaysia

    arithmetic progression

    A training programme requires a footballer to run circuits of a football field every day . On each day , he runs three circuit more than he did the day before . On the fifth day he runs 14 circuits . Calculate the number of circuits covered by him altogether from the end of the n th day to the end of the ( 2n ) th dat , in terms of n

    Can someone pls explain to me .Thank s
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  2. #2
    Junior Member
    Oct 2008
    Term n of an arithmetic series given by
    Tn= a + (n-1)d

    where a = first term and d = common difference
    d is given = 3
    T5 = a + (5-1)3 = a + 12 = 14
    so a = 2

    Sn the summation of the arithmetic series is given by

    Sn= n/2[2a +(n-1)d]

    so in this case Sn= n/2[4+3(n-1)]

    and S2n= 2n/2[4+3(2n-1)]

    S2n - Sn reduces to n/2(9n+1)
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